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Proof step 2 from proof restn-first-get (previous step)

∀ n. n < # x0 → restn(n, x0).first = x0[n] ⊦ n < #(a + x0) → restn(n, a + x0).first = (a + x0)[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n ≤ # x0, ∀ n. n < # x0 → restn(n, x0).first = x0[n] ⊦ restn(n, a ' + x0).first = (a ' + x0)[n]

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list